The rates of a reaction starting with intial conentrations of 2xx10^(-3) M and 1xx10^(-3) M are equal to 2.40xx10^(-4) M s^(-1) and 0.60xx10^(-4) M s^(-1) respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.

The rates of a reaction starting with intial conentrations of 2xx10^(-

1.	The rates of a reaction starting with intial conentrations of 2xx10^(-3) M and 1xx10^(-3) M are equal to 2.40xx10^(-4) M s^(-1) and 0.60xx10^(-4) M s^(-1) respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.
Answer» Solution :If N is the order of reaction RATE `=k[A_(0)]^(n)=kC^(n)` For two different initialconcentrations, we have `(r_(1))/(r_(2))=((C_(1))/(C_(2)))^(n):.log""(r_(1))/(r_(2))=nlog""(C_(1))/(C_(2))` or `""n=(log(r_(1)//r_(2)))/(log(C_(1)//C_(2)))=(log(2.4xx10^(-4)//0.6xx10^(-4)))/(log(2xx10^(-3)//1xx10^(-3)))=(LOG4)/(log2)=2` Hence, orderof reaction = 2 Rate `=k[A_(0)]^(2)` `k=("Rate")/([A_(0)]^(2))=(2.4xx10^(-4)"mol L"^(-1)s^(-1))/((2xx10^(-3)"mol L"^(-1))^(2))=0.6xx10^(2)"L mil"^(-1)s^(-1)`